# When does the product of two polynomials = $x^{k}$?

Suppose $f$ and $g$ are are two polynomials with complex coefficents (i.e $f,g \in \mathbb{C}[x]$). Let $m$ be the order of $f$ and let $n$ be the order of $g$.

Are there some general conditions where

$fg= \alpha x^{n+m}$

for some non-zero $\alpha \in \mathbb{C}$

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Unless f and g are monomials themselves, I can't see how else this will hold. –  Ｊ. Ｍ. Aug 26 '10 at 14:51
TIP It's not a good idea to accept an answer 10 minutes after posting your question. Being accepted, this will cause the software to give the post much less exposure, so you will have much less a chance to receive other good answers. And, generally, it's highly unlikely that you'll get the best answer in 10 minutes. Generally you shouldn't accept an answer for at least a few days if not more if you want to have the best possible chance of learning from the collective expertise here. –  Bill Dubuque Aug 26 '10 at 15:09
Sorry I will keep that in mind. I was somewhat eager to close the question since the answer occured to me a few minutes after I posted it. –  Digital Gal Aug 26 '10 at 15:23

Polynomials over $\mathbb{C}$ (in fact, over any field) are a Unique Factorization Domain (see http://en.wikipedia.org/wiki/Unique_factorization_domain); since $x$ is an irreducible, the only way for that to happen is for $f=ax^m$ and $g=bx^n$, with $ab=\alpha$.

(If you don't want to bring in the sledgehammer of unique factorization, you can just do it explicitly: look at the lowest nonzero term in $f$ and the lowest nonzero term in $g$; their product will be the lowest nonzero term in $fg$, hence must be of degree $m+n$. Since the degree of the lowest nonzero term of $f$ is at most $m$ and the one of $g$ is at most $n$, you have that they must be exactly of degree $m$ and $n$, respectively, and you get the result)

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Here one needs only the weak statement that products of primes factor uniquely - not the strong statement that products of irreducibles factor uniquely ($\Rightarrow$ UFD when factorizations are finite). But this weaker statement is true in any domain. Indeed, the well-known one-line proof for $\mathbb Z$ generalizes immediately. That $x$ is prime here is immediate - see my answer. –  Bill Dubuque Aug 29 '10 at 17:58

It's true over any domain $\rm D$ since then $x$ is prime (via $\rm D[x]/x = D$ a domain), and products of primes always factor uniquely in domains. It fails over non-domains, e.g. $\rm x = (2x+3)(3x+2) \in \mathbb Z/6[x].$

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The answer just occured to me. The roots of $f$ and $g$ must be be at 0.

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That is a good way of seeing things for complex polynomials. ^^ +1 –  Patrick Da Silva Aug 29 '11 at 1:00

Yes, the intuitively evident ones: all other terms in f and g must vanish. To see this, note that the product of the constant terms of f and g equals the constant term of fg, which is zero, whence at least one of these polynomials is multiple of x. Without any loss of generality assume it is f. Then

fg = x (f/x) g,

implying (f/x) g is a multiple of x^(n+m-1). By induction this reduces us to the case n+m=0, which is trivial (because f and g then have no other terms). QED.

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